Charles’ Law
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1 Charles’ Law • At constant pressure, the volume of a gas varies directly with its Kelvin temperature • Equations: V 1 V 2 V 2 V 1 = T 2 T 1 = T 1 T 2 or or V 1 T 2 = V 2 T 1
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Charles’ Law. At constant pressure, the volume of a gas varies directly with its Kelvin temperature Equations: V 1 V 2 V 2. T 2. =. =. V 1 T 2 = V 2 T 1. or. or. T 1. T 2. V 1. T 1. At Constant Pressure,. when volume , temp When volume , temp. - PowerPoint PPT Presentation
Transcript of Charles’ Law
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Charles’ Law
• At constant pressure, the volume of a gas varies directly with its Kelvin temperature
• Equations:
V1 V2 V2
V1
=T2
T1
=T1 T2
or or V1T2 = V2T1
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At Constant Pressure,
• when volume , temp
• When volume , temp
Temp. scale K not °CWhat error exists
on this graph?
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Solving Charles’ Law Problems
– Write the given information– Convert to Kelvin temp. scale– Isolate the missing variable – Plug-in the given values and solve
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Example:A volume of 20.0m3 of argon gas is kept under constant
pressure. The gas is heated from 22ºC to 283 ºC. What is the new volume of the gas?
Step 1: Write down the information you know
V2 = ?
V1 = 20.0 m3
T1 = 22ºC
T2 = 283 º C
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Step2) Convert Temperatures into KelvinoC + 273
T1= 22oC 22 + 273 = 296
T2= 283oC 283 + 273 = 556K
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Step 3) Isolate the unknown variable
V1T2 = V2T1 We don’t know V2
Divide both sides by T1
V1T2 = V2T1
T1 T1
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Step 4) Plug in the numbers you have and solve
(20.0 m3)(556 K) = V2 (Note: Kelvin
296 K units cancel)
37.6 m3 = V2
